prove this identity $\sin(x+y)\sin(x-y)=\sin^2 x - \sin^2 y$

Question

prove this identity : $$\sin(x+y)\sin(x-y)=\sin^2 x - \sin^2 y$$ I tried solving it with additional formulas but I can't get the right answer. I get $$\sin^2 x \cos^2 y-\cos^2 x \sin^2 y$$

Do you want a geometrical proof or just a proof using some common facts like $\sin^2+\cos^2=1$? — Ragnar, Feb 06 '14 at 09:27

score 4 · Accepted Answer · edited Feb 06 '14 at 09:37

4

$\sin^2 x \cos^2 y-\cos^2 x \sin^2 y=\sin^2 x(1-\sin^2 y) -(1-\sin^2 x) \sin ^2 y$

$=\sin^2 x -\sin^2 x\sin^2y -\sin ^2 y + \sin^2x\sin^2y$

$=\sin^2 x - \sin ^2 y$

edited Feb 06 '14 at 09:37

lsp

4,745

answered Feb 06 '14 at 09:30

kmitov

4,731

score 1 · Answer 2 · answered Feb 06 '14 at 09:30

Use the identity $$\sin(x\pm y)=\sin x\cos y\pm \sin y\cos x$$ and we can get \begin{align*} LHS&=\sin(x+y)\sin(x-y)\\ &=(\sin x\cos y+\cos x\sin y)(\sin x\cos y-\cos x\sin y)\\ &=\sin^2x\cos^2y-\cos^2x\sin^2y \\ \end{align*}

See if you can take it from here using the identity $$\sin^2x+\cos^2x=1.$$

prove this identity $\sin(x+y)\sin(x-y)=\sin^2 x - \sin^2 y$

2 Answers2